Contrasting 1×2×3... with 1+2+3... I was wondering what comparing the graph of 1+2+3... with 1×2×3... would look like. In doing so, I tried graphing 1×2×3... in Excel, but no matter how many points I plotted (e.g. 5, 120 vs. 8, 40320) I kept getting a jagged angle towards the end, like so:

where $y\in 1,2,6,24,120,720,5040,40320$.
Compare this with the smooth

Is the graph just a series of successive jagged angles? Also, what is the derivative of the series 1×2×3...? How much faster does it grow than 1+2+3...? What are the two series called? I can't seem to find a graphing software online whihch will allow me to scale down the x asis and scale up the y axis so that I can see how the two functions grow in the same amount of time. Such a graph would be appreciated.
 A: Here are the graphs of the functions $f(n)=1+2+\cdots+n$ (blue) and $g(n)=1\times2\times\cdots\times n$ (orange) for $n=5$, $n=10$ and $n=100$
$\mathbf{n=5}$

$\mathbf{n=10}$

$\mathbf{n=100}$ (here the blue graph for addition is barely visible)

The blue function appears smooth, for example as below for $n=100$.

However it is not, it has a jagged angle at every integer value on the $x$-axis. The reason that it appears smooth is that the change of angle at $x=n$ and $x=n+1$ is smaller and smaller as $n$ grows larger: the slope of the line segment between the points $(n,f(n))$ and $(n+1,f(n+1))$ is
$$\frac{f(n+1)-f(n)}{n+1-n}=\frac{(1+2+\cdots+n+1)-(1+2+\cdots+n)}{1}=n+1,$$
so the ratio of the slopes of two consecutive line segments is $\frac{n+1}{n}=1+\frac{1}{n}$. Therefore, as $n$ grows larger and larger, the slopes of two consecutive line segments change by only a very small amount that eventually the human eye can't discern and because of that the graph appears smooth.
Contrast this with the orange graph for the product: first, we write $1\times2\times\cdots\times n$ as $n!$ (read as '$n$ factorial'). The slope of the line segment between the points $(n,g(n))$ and $(n+1,g(n+1))$ is
$$\frac{g(n+1)-g(n)}{n+1-n}=\frac{(n+1)!-n!
}{1}=(n+1)!-n!=n!((n+1)-1)=n!\cdot n.$$
The ratio of the slopes of two consecutive line segments is now
$$\frac{(n+1)!\cdot (n+1)}{n!\cdot n}=\frac{(n+1)(n+1)}{n}=\frac{n^2+2n+1}{n}=n+2+\frac{1}{n}.$$
Therefore, as $n$ grows larger and larger, the slopes of two consecutive line segments change by a huge amount, and the non-smoothness is more apparent because the human eye can easily distinguish between such a drastic change in angles.
